If -2 and 3 are solutions of the quadratic equation x^2 + bx +c = 0, then what are the values of b and c?

Method 1 :
Sum of the roots :
(-2) + 3 = -b
b = -1

Product of the roots :
(-2) × 3 = c
c = -6


Method 2 :
x = -2 or x = 3
(x + 2) = 0 or (x - 3) = 0
(x + 2)(x - 3) = 0
x² - x - 6 = 0

Compare the equation (x² - x - 6 = 0) with the equation (x² + bx + c = 0)
b = -1 and c = -6


Method 3 :
Sub. x = -2 into the given equation :
(-2)² + (-2)b + c = 0
-2b + c = -4 ...... [1]

Sub. x = 3 into the given equation :
(3)² + (3)b + c = 0
3b + c = -9 ...... [2]

[2] - [1] :
5b = -5
b = -1

Sub. b = -1 into [1] :
-2(-1) + c = -4
c = -6

If -2 and 3 are solutions, then:
x = -2
(x+2)=0
and liewise
(x-3) = 0

together we have (x+2)(x-3) = 0
Multiply the terms using FOIL:
x² -3x +2x -6 = 0
combine like items:
x² -x -6 = 0
b= -1
c= -6

x^2 + bx + c = 0. (x+2)(x-3)=0. x^2 -x -6 = 0. (b,c) = (-1,-6).